Answer:
1) acceleration is increased by a factor of four 4X
2) the acceleration increases a factor of 2X
3) the correct answer of 400g
Explanation:
This is a kinematics exercise, where you use the velocity equation to obtain the acceleration, with the final velocity equal to zero.
v² = v₀² + 2 a x
0 = v₀² + 2 a x
a = - v₀² / 2 x
In the case of wanting to give the acceleration as a function of g, we can find the relationship between the two quantities
a / g = - v₀² / (2 x g)
Let's answer the different questions about this equation
1. The initial velocity is doubled, how much the acceleration is worth
a/g = - (2v₀) 2 / 2xg
a = 4 (-v₀² / 2xg) g
acceleration is increased by a factor of four 4X
2. if the stopping distance is reduced by 2, that is, x = x₀ / 2
we substitute
a/g = (- v₀² / 2g) 2/x
a =2 (-v₀² / 2x₀g) g
therefore the acceleration increases a factor of 2X
3. the initial velocity of the hockey player is v₀ = 20 m / s and the stopping distance is
x = 5cm = 0.05m
we calculate the acceleration
a / g = - 20² / (2 0.05)
a / g = - 4000 / g
a / g = - 4000 / 9.8 = 408
a = 408 g
the correct answer of 400g, the value matches exactly if g = 10 m / s2 is taken
